Do integrable functions decay to zero?

**aukie** · December 17th 2010, 07:38 AM

Hello

I've heard if $f \in L^1(\mathbb{R})$ , the space of absolutley integrable fuctions, then f must decay to zero, i.e. $\lim_{x\to \infty } f(x)=0$ . Is this true? Can anyone direct me to a proof (book or online)? Is there a more general result regarding the asymptotics of integrable functions?

Thanks

**Ackbeet** · December 17th 2010, 07:43 AM

Not true unless the function is also differentiable or some smoothness condition applied (probably Lipschitz is sufficient). Counterexample:

$f(x)=\begin{cases}e^{-x^{2}}\quad& x\in\mathbb{Q}^{c}\\<br /> 1\quad& x\in\mathbb{Q}\end{cases}.$

Definitely an absolutely (Lebesgue) integrable function, with integral $\sqrt{\pi},$ in fact, but the limit in question does not exist.

Now, you might be able to say that the lim sup has to go to zero. Haven't examined that in detail.

**chisigma** · December 17th 2010, 08:07 AM

Originally Posted by aukie

Hello

I've heard if $f \in L^1(\mathbb{R})$ , the space of absolutley integrable fuctions, then f must decay to zero, i.e. $\lim_{x\to \infty } f(x)=0$ . Is this true? Can anyone direct me to a proof (book or online)? Is there a more general result regarding the asymptotics of integrable functions?

Thanks

An interesting 'counterexample' is the integral...

$\displaystyle \int_{-\infty}^{+ \infty} \cos x^{2}\ dx = \sqrt {\frac{\pi}{2}}$ (1)

... where the $\lim_{x \righatarrow \pm \infty} f(x)$ doesn't exists...

Merry Christmas from Italy

$\chi$ $\sigma$

**Ackbeet** · December 17th 2010, 09:04 AM

It is an interesting example. Is your integrand absolutely integrable?

**chisigma** · December 17th 2010, 11:24 AM

The general case of this type of integral has been examined in...

http://www.mathhelpforum.com/math-he...ls-157194.html

Merry Christmas from Italy

$\chi$ $\sigma$

**Ackbeet** · December 17th 2010, 05:29 PM

Hmm. Not sure I see how that shows that

$\displaystyle \int_{-\infty}^{+ \infty}|\cos(x^{2})|\,dx<\infty,$ which is what it means for the integrand to be absolutely integrable. What do you think?

**Drexel28** · December 17th 2010, 08:52 PM

Originally Posted by Ackbeet

Hmm. Not sure I see how that shows that

$\displaystyle \int_{-\infty}^{+ \infty}|\cos(x^{2})|\,dx<\infty,$ which is what it means for the integrand to be absolutely integrable. What do you think?

In fact, this isn't true, as you guessed this integral diverges. The proof is similar to why $\displaystyle \int_0^{\infty}\frac{|\sin(x)|}{\sqrt{x}}\text{ }dx$ diverges. Ask for a proof if you'd like one.

**chisigma** · December 18th 2010, 12:14 AM

Honestly I hardly undestand the scope to distinguish between a function for which the integral...

$\displaystyle \int_{- \infty}^{+ \infty} f(t)\ dt$ (1)

... symply 'exists' from a function for which the integral (1) exists but the integral...

$\displaystyle \int_{- \infty}^{+ \infty} |f(t)|\ dt$ (2)

... doesn't... as well as the utility of the so called 'phantom functions' [i.e. a function that is 'almost everywhere' equal to zero]...

... anyway if You all desire a function $f(t) \in L^{1}$ that contains no 'phantom fuction' and that han no limit for $t \rightarrow \infty$ I propose this one...

$\displaystyle f(t)=\left\{\begin{array}{ll}1 ,\,\,n-\frac{1}{2\ (n+1)^{2}} < t < n + \frac{1}{2\ (n+1)^{2}}, n \in \mathbb{N}\\{}\\0 ,\,\, elsewhere \end{array}\right.$ (3)

Of course is $f(t) \ge 0$ , there is no limit for $t \rightarrow \infty$ and is...

$\displaystyle \int_{- \infty}^{+ \infty} f(t)\ dt = \frac{\pi^{2}}{6} - 1$ (4)

Merry Christmas from Italy

$\chi$ $\sigma$

**Opalg** · December 18th 2010, 12:16 AM

Originally Posted by aukie

Hello

I've heard if $f \in L^1(\mathbb{R})$ , the space of absolutely integrable functions, then f must decay to zero, i.e. $\lim_{x\to \infty } f(x)=0$ . Is this true? Can anyone direct me to a proof (book or online)? Is there a more general result regarding the asymptotics of integrable functions?

That is not true, even for continuous absolutely integrable functions. A counterexample would be a function that has a succession of narrow spikes, such as

$f(x) = \left\{\begin{array}{l}\;\;0\hspace{6.2em}(x<2)\\ \left.\begin{array}{ll}n^2(x-n)&(n\leqslant x < n+\frac1{n^2})\\2 - n^2(x-n)&( n+\frac1{n^2}\leqslant x < n+\frac2{n^2})\vspace{0.5ex}\\ 0&(n+\frac2{n^2}\leqslant x< n+1) \end{array}\right\}\quad(n\geqslant 2).\end{array}\right.$

The integral of $f$ is the sum of the areas under the spiky triangles, namely $\sum_{n=2}^\infty\frac1{n^2} < \infty$ .

With a little care, you could even smooth out the spikes to get a differentiable example.

Edit. Just beaten to it by chisigma.

**Bruno J.** · December 18th 2010, 02:40 PM

Originally Posted by chisigma

Honestly I hardly undestand the scope to distinguish between a function for which the integral...

$\displaystyle \int_{- \infty}^{+ \infty} f(t)\ dt$ (1)

... symply 'exists' from a function for which the integral (1) exists but the integral...

$\displaystyle \int_{- \infty}^{+ \infty} |f(t)|\ dt$ (2)

... doesn't... as well as the utility of the so called 'phantom functions' [i.e. a function that is 'almost everywhere' equal to zero]...

The difference is that the Lebesgue integral (1) does not exist unless (2) is satisfied.

As for functions which behave in a certain manner almost everywhere (i.e. everywhere but on a set of zero measure), they are crucial in providing counterexamples to many intuitive misconceptions about the Lebesgue integral, which originate from our intuition of the Riemann integral. Also, many theorems of analysis can only assert properties of functions which hold almost everywhere. (For example, take Lebesgue's theorem, which states that an increasing function on a closed, bounded interval is almost everywhere differentiable.)

**chisigma** · December 19th 2010, 12:11 AM

Originally Posted by Bruno J.

The difference is that the Lebesgue integral (1) does not exist unless (2) is satisfied.

As for functions which behave in a certain manner almost everywhere (i.e. everywhere but on a set of zero measure), they are crucial in providing counterexamples to many intuitive misconceptions about the Lebesgue integral, which originate from our intuition of the Riemann integral. Also, many theorems of analysis can only assert properties of functions which hold almost everywhere. (For example, take Lebesgue's theorem, which states that an increasing function on a closed, bounded interval is almost everywhere differentiable.)

That's [among many other arguments...] a good argument to discuss in the 'freely math forum' You have proposed... Your idea is very good!...

Merry Christmas from Italy

$\chi$ $\sigma$

**chisigma** · December 19th 2010, 01:39 AM

May be it would be useful to spend some words about the function 'duty modulated pulse train'...

$\displaystyle f(t)=\left\{\begin{array}{ll}1 ,\,\,n-\frac{1}{2\ n^{2}} < t < n + \frac{1}{2\ n^{2}}, n \in \mathbb{N}\\{}\\0 ,\,\, elsewhere \end{array}\right.$ (1)

The L-transform of (1) is...

$\displaystyle \mathcal{L} \{f(t)\} = \frac{1}{s} \{\sum_{n=1}^{\infty} e^{-s\ (n-\frac{1}{2\ n^{2}})} - \sum_{n=1}^{\infty} e^{-s\ (n+\frac{1}{2\ n^{2}})} \} = \frac{2}{s} \sum_{n=1}^{\infty} e^{-s\ n}\ \sinh \frac{s}{2\ n^{2}}$ (2)

... so that the L-tranform of integral of (1) is...

$\displaystyle \mathcal{L} \{\int_{0}^{t} f(\tau)\ d \tau\} = \frac{1}{s^{2}} \{\sum_{n=1}^{\infty} e^{-s\ (n-\frac{1}{2\ n^{2}})} - \sum_{n=1}^{\infty} e^{-s\ (n+\frac{1}{2\ n^{2}})} \} = \frac{2}{s^{2}} \sum_{n=1}^{\infty} e^{-s\ n}\ \sinh \frac{s}{2\ n^{2}}$ (3)

Now if we apply the so called 'final value theorem' to (3) [l'Hopital's rule...] we obtain...

$\displaystyle \lim_{t \rightarrow \infty} \int_{0}^{t} f(\tau)\ d \tau = \sum_{n=1}^{\infty} \frac{1}{n^{2}}$ (4)

... and that is not a 'surprise'. A little 'surprise' however happens if we [try to] apply the 'final value theorem' to (2) obtaining...

$\displaystyle \lim_{t \rightarrow \infty} f(t) = 0$ (5)

... that contradicts the fact that the limit if t tends to infinity of f(t) doesn't exist... a spontaneous question: what is my 'mistake'?...

Merry Christmas from Italy

$\chi$ $\sigma$

**aukie** · December 19th 2010, 05:52 AM

Cheers very easy to understand example.

**aukie** · December 19th 2010, 06:21 AM

Thank you Ackbeet for providing a counter example (that is almost everywhere continuous)

Thank you chisigma for providing a counter example (that is piecewise continuous)

and Thank you Opalg for providing a counter example (that is continuous)

I am amazed at how all of you have conjured up these examples (especially the last two). Since I am trying to learn math on my own can I ask a rather practical question, When faced with such complicated functions, is it best to use mathematica or maple to understand these functions or should you just try and spend some time with pen and paper? I was just wondering what real mathematicians do.

These counterexamples are certainly unintuitive (at least for me. I would have thought the function would decay to zero).

The examples also leave one more question raised in my mind. The reason I asked it is because I was trying to work through a proof in Folland's book on Fourier analysis (Theorem 7.5c).

In his proof the conditions he gives on the function f are as follows:

f is continuous
f is piecwise smooth (i.e. f' is piecewise continuous)
$f \in L^1(\mathbb{R})$
$f' \in L^1(\mathbb{R})$

Then he writes, "Observe that since $f' \in L^1(\mathbb{R})$ , the limit,

$\lim_{x\to \infty } f(x)=f(0)+ \int _0^{\infty }f'(x)dx$

exists, (I get this, simple application of the fundamental theorem of calculus),

and since $f \in L^1(\mathbb{R})$ this limit must be zero (this part, I dont get!)"

Any help with this would be appreciated. From the counter examples provided I can see the only extra condition on f is that it is piecewise smooth. Would this then make $f$ tend to zero as $x$ goes to infinity? I couldnt quite prove this.

I guess not since Opalg example is piecewise smooth! I just dont see how the statement is true then??

Incidentally the theorem states:

Suppose $f \in L^1(\mathbb{R})$ , if f is continuous and piecewise smooth and $f' \in L^1(\mathbb{R})$ , then

$\hat{[f']}(\xi) = i \xi \hat{f}(\xi)$ ,

where $\hat{f}(\xi)$ is the fouries transform of f and $\hat{[f']}(\xi)$ is the fourier transform of $f'$

**Opalg** · December 19th 2010, 06:42 AM

Originally Posted by aukie

I was trying to work through a proof in Folland's book on Fourier analysis (Theorem 7.5c).

In his proof the conditions he gives on the function f are as follows:

f is continuous
f is piecwise smooth (i.e. f' is piecewise continuous)
$f \in L^1(\mathbb{R})$
$f' \in L^1(\mathbb{R})$

Then he writes, "Observe that since $f' \in L^1(\mathbb{R})$ , the limit,

$\lim_{x\to \infty } f(x)=f(0)+ \int _0^{\infty }f'(x)dx$

exists, (I get this, simple application of the fundamental theorem of calculus),

and since $f \in L^1(\mathbb{R})$ this limit must be zero (this part, I dont get!)"

The crucial extra piece of information here is that $f' \in L^1(\mathbb{R})$ and hence the limit $\lim_{x\to \infty } f(x)$ exists. If the limit exists, and the function is in $L^1(\mathbb{R})$ , then the limit must be zero. (If the limit is not zero, then the integral of $|f|$ could not be finite.)

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